# Find n if 2^200 - 31.2^192 + 2^n is a perfect square

Most challenging question of my life,I want to share this with my friends..enjoy...please,this is not my homework..

Note by Patel Kishan
5 years, 7 months ago

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So $$n = \boxed{198}$$ is the only solution to the problem

- 5 years, 7 months ago

Here is the full solution to this problem.

i) if $$n = 192$$,

Then we have $$2^{192}(226)$$, which is clearly not a perfect square.

ii) if $$n > 192$$,

Then since $$2^{192}$$ is a perfect square, and every term is greater than $$2^{192}$$, dividing by $$2^{192}$$ yields

$$225 + 2^{n - 192}$$, which must also be a perfect square.

Let $$225 + 2^{n - 192} = k^{2}$$,

Then $$2^{n - 192} = (k + 15)(k - 15)$$.

Note that both factors must be powers of two and their difference is $$30$$.

Note that $$2^{6} - 2^{5} = 2^{5} > 30$$, so both factors must be less than $$2^{6}$$

By considering all cases, we get only $$30 = 2^{5} - 2^{1}$$ works.

And so we get $$k + 15 = 32, k - 15 = 2$$, so $$2^{n - 192} = 32\cdot 2 = 2^{6}$$ and thus $$n = 198$$.

iii) if $$n < 192$$

$$2^{200} - 31\cdot 2^{192} + 2^{n} = 2^{n}(2^{200 - n} - 31\cdot 2^{192 - n} + 1)$$

Note that the factor in the bracket is odd

And so if $$n$$ is odd, then $$2^{n}$$ is not a perfect square and so the factor in the bracket must be even, contradiction.

So $$n$$ is even.

Then $$2^{n}$$ is a perfect square.

So $$2^{200 - n} - 31\cdot 2^{192 - n} + 1$$ is a perfect square.

$$2^{200 - n} - 31\cdot 2^{192 - n} + 1 = (2^{192 - n}(15^{2})) + 1$$

Since $$n$$ is even, $$192 - n$$ is also even.

So $$2^{192 - n}$$ is a perfect square.

And so $$2^{200 - n} - 31\cdot 2^{192 - n} + 1$$ is not a perfect square since it is one more than a perfect square.

Note : $$2^{192 - n} \ne 0$$

- 5 years, 7 months ago

n = 198

- 5 years, 7 months ago

@Samuel, the reason is because you entered the brackets as \ [ code \ ], instead of \ ( code \ ) . The former will make each equation into a line, the latter will make each equation part of your sentence. If you look at the math formatting guide, all of the brackets are \ ( \ ). This is similar to the difference between  code , and $code$ in other Latex compilers. (spaces suitably removed).

Staff - 5 years, 7 months ago

Zi Song, that's correct, albeit extremely lengthy. Can you shorten it to just a few lines? I believe that you all you need is $$225 \times 2^{192} + 2^n = k^2$$, and that there is a unique solution $$(a,b)$$ to $$2^a - 2^b = c$$, for any integer $$c$$. You explicitly used the first idea, and implicitly used the second idea in "By considering all cases ...".

Staff - 5 years, 7 months ago

Your blog is awesome,why do you stop writing it,please if you get some good questions,then please send it to me...by the way, how did you become so good in maths.....please share it with me pleaseeeee

- 5 years, 7 months ago

But however i think Brilliant scholars blog is much much better than any blog ! So recommend you to read posts from brilliant scholars blog.

- 5 years, 7 months ago

- 5 years, 7 months ago

Sorry i am weak at maths,please tell me Then since 2^192 is a perfect square, and every term is greater than 2^192 , dividing by yields.......why you only divide the whole term by 2^192...........? and why divide only..... please help me friends.....

- 5 years, 7 months ago

Naishad P. - $$n = 198$$ is indeed correct. Because $$2^{192}(2^8 - 31 + 2^6) = 2^{192}\cdot 17^2$$ is a perfect square.

- 5 years, 7 months ago

You are saying $$n=196$$ then it that case

$$2^{192}(2^8 - 31 + 2^4)$$ since $$2^{192}$$ is a square , $$256 - 31 +16$$ should be s square but since it is not a perfect square , this means that $$n=196$$ is not the answer :

now consider three cases : $$n>192 ,n= 192 , n < 192$$ and do case chase you will land up with

$$\boxed{n=198}$$ is the only solution

- 5 years, 7 months ago

friends i hope the answer would be 196...is it correct,i dont have its answer.....

- 5 years, 7 months ago

So since it's not specified that the solution has to be an integer, I came up with this:

[note: let me apologize ahead of time for the formatting of this comment. For some reason the $\LaTeX$ processor on Brilliant is giving all of my math a new line (as above).]

$2^{200}-31 \times 2^{192}+2^{n}=$ $2^{184}(2^{16}-31 \times 2^{8}+2^{n-184})$

Letting $a=2^{8}$ we have

$2^{184}(a^{2}-31 \times a+2^{n-184})$

Since 2^184 is a perfect square, all we need to show is that $a^{2}-31 \times a+2^{n-184}$ is a square. However, note that it does not need to be the square of an integer; it suffices to be an integer divided by an even power of 2 less than or equal to 184 (I know this is wordy. Don't think about it too much now because it will make sense in context later.)

We now use the identity that $(a-c)^{2}=a^{2}-2ac+c^2$. Therefore when $2^{n-184}=(\frac{31}{2})^{2}$, our quantity is a square. Now, we only need to solve a one-variable equation:

$2^{n-184}=(\frac{31}{2})^{2}$ $2^{2}2^{n-184}=931$ $2^{n-182}=931$ $n-182=\log_2 931$ $n=log_2 931+182$

This makes $2^{16}-31 \cdot 2^{8}+2^{n-184}=\frac{231361}{4}$ Alas! We've made the square of $2^{8}-\frac{31}{2}=\frac{481}{2}$ which is not an integer! But wait! We get lucky! When multiplying it by 2^184, it corrects, and we get an integer with an integral square root!

And indeed, according to my calculator $\sqrt{2^{200}-31 \times 2^{192}+2^{n}}$ for $n=\log_2 931+182$ is equal to 1190898317792535824452957503488, an integer!!!!! Thus, we are done.

- 5 years, 7 months ago